In $60$ minute the hour hand moves 30 level so in 35 minutes it would have actually moved

$$\frac3060 \times 35$$$$=17.5^\circ$$

Hour hand in ~ $17.5^\circ$ native 7 . Minute hand exactly points at 7 , so angle in between minute hand and also hour hand = $17.5^\circ$

I would allude you come an digital tool

clock edge calculator

Hint: Measuring angle clockwise from directly up in degrees, at $7:00$ the minute hand is at $0^\circ$ and the hour hand is in ~ $210^\circ$. What is the price of motion of the hour hand in degrees/hour? What fraction of an hour is $35$ minutes? The minute hand will certainly be in ~ $210^\circ$

$1$ hour because that the hour hand = $5$ minutes because that the minute hand = 30°.

You are watching: 7:35 on a clock

The minute hand is obviously in ~ $210°$.

The hour hand is between $210°$ and also $240°$. Now, to gain the exact amount in between this two:$$x:35 = 30°:60$$$$x:35 = 1°/2$$$$x = 35°/2 = 17.5°$$

Hint: because 35 minutes have actually elapsed out of the 60 minute in one hour, the hour hand will have moved $\frac3560$ that the angle between the 7 and the 8. Now, what is the angle between the 7 and also the 8?

The minute hand will have travelled $\frac3560\cdot 360$ levels from "straight up."

The hour hand is a small trickier. Every hour it travels with $\frac36012=30$ degrees. So in ~ $7:35$ it has travelled $\left(7+\frac3560\right)\cdot 30$ from right up.

Calculate, subtract.

assuming the hour hand moves "continuously over time" then the edge the hour hand indigenous 12 o"clock is $2\pi(7+ \frac3560)/12$ and also the angle of the minute hand is $2\pi(35/60)$... So just find the difference.

assuming:h: 0-11 -> hourm: 0-59 -> minutes: 0-59 -> secondthen

the always correct formula is:A = 30 h - 11/2 m - 11/120 sAngle = min

and in your case:h = 7m = 35s = 0=> A = 30*7 - 11/2 * 35 = 17.5Angle = min , 360 - = min 17.5 , 342.5 = 17.5

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